prob.xi-8n - 32.031 AND 22.312 ENGINEERING OF NUCLEAR...

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Unformatted text preview: 32.031 AND 22.312 ENGINEERING OF NUCLEAR REACTDRS I -_--'-_ PROBLEM I I-SN HEM Pressure L355 PranEm angidm a 3 mam, lung “filler Shanna] Drum-1m” CTflEfi-Smlional ma 1.5:IEIIII""*IE2 flpcraiing at the bllcrwing conditions: T'Il = [1.29 1(ng p = 37.2 Ma mflpmfi Hm Prawn": has um” humogflnmm “finiiibfium ELSE'-”1-'lF*1i1::ns for [he fallnwing additinna] Emldiljuns: - a: Adiabatic channel with inlet flow quality of I]. IS. .. h. Unifnnn axial heat flux Ef sufficient magnitude to heal. the entering saturamd cuulant In 2111 exit qualily' UF [1.] 5. {E 1E»: r: HHS. ' L .E' III 'I .— 4-..- ; “:r ".T . PmQO 11-53 ‘w' Given Data: m =0.29 kgfis, A =15 x If)" r112 xni: = [115. P = T2 MP3 L = 3 :11 Saturated thermal properties @ P=l2 MPa pr 2 736.49 kgfm3+ fig = 31?] kgfm3 u = 96.931: 10*5 Pa-s Tetal pressure drop: 45PM: = flPeer: + Jungva + fitPfriclinn + flpferm Since pipe has eenslunt area, there is no fer less. fiPm: = er + eP gra‘flly + flPI'IICLJUn GEJ (6*) &P 2 _EL _ -_m._ am: ([04:13 But P31 in L fiPErHviry = 1. ngdz El L {G (3.. 3 . fiPfriaition : m m] $|Q{Z} :12 II} 2D'elz‘l .- . ' where. 2.: "_- meZ} = J ' UWJWI ' Pei . " .I: __1__ =f_{1-x{z}]: + [3(3)]: L :2sz _{1 — {xinin 012mg; -'-_ {930(2) = ‘P—r - I]x[z_} + I ‘-' - -._:;:.:" F3: {-5 . Using HEM. 5 =1 mEz} = 1 {13.44} (1) {13-451)} {'2} (3} (11-32] r m 11-3 ntin ed Gm = u'l = 42—9—— =1933.33 13% r 1.5 x 10'4 1113-5 I "11- Dfl=4fé$=qI4—______W Re: GED: = {1933.33}?0.9138} = 2352x105 H 95.933103 -3_ McAdame' Correlatien ' r: 0.1341232 = 0.134(3752x 1051‘“ = 0.015 Afliabatic Channel Tilz'Sinee x65} is cansteutfflJS}, “[33 is also constant. fl=———___1_____. [1+{1- HELMJ 0.15 736.49 = 0.??5 . at?“ = [1 .‘"'ii) Gravity pressure drop: : fiPgm-n'ry = ngL = m “WWI -pg}J3L = 335.49 - 0333335349 - 3?.71}]9.3 3-. 3 = 5323 Pa = 5.73 km iii) Frietien pressure drUP‘ Ifi'l-Pfri-ztievnl = RI: - 1’}: + IJL E ..,..__ .- - = 9'.015"[1——__..933'33J [Fifi—9.4 _ 1]0.15 + H1313 _':::.:; 2{U.01331T35.49} 37.71 ' = 313301311 = 31.33 kPa Therefore. 51PM = 5.73 + 31.33 = 37.06 ;.i) Acceleratifln pressura (imp: ‘95me = I [l'xcxitF + {XMELFI _ Lia; — ———-———__. + l {1 ~ u.’:75}73-5.49 0.7?5x3731 = 141mm = 14.1]kPa ii) vaity pressure Limp: L anm-w =f ha: a MIND] - Pgflgdz CI 34 em {1+{l-xJEE X Pr flpgravity : [pi ' wk}! 0 L flP raiij = Er—EL I - ME 1;: 3“ xuirf [pf+ {1 . :5] Daddy: L . {1 fi “3t” 1315;!“ jdx cm a “31" Pfl'x + P}; = REE: [ --———3—p ——]dx Jlithlu'l u {pi _ 135+): + pg [H.155 _1_ iii} Friction pressure drop: L fiqumer _ flip ' n = " 1 1 d2 _ frtcun 20cm faflpg FLU '1' I = —————fi”Gmp"‘I - 1J1: + 1} 31—34311 2'3th Pg :12»: = made-14.; “fl ~ le + IJ dx EDepi chil a P1: = _ + flak] 2133;); N"mill Pg 2 0.01541933332 3 [116$ 2 j =——-*——-———— -1}-G—Ji+0.153.0 2(0.0133[?36.49)0.1S{3?.7‘1 2 = 19811} PH 219.31kPa :- Therefore. flPm = 14.11 + 10.36 + 19.31 = 44.28 kPa ...
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prob.xi-8n - 32.031 AND 22.312 ENGINEERING OF NUCLEAR...

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